[LLVMdev] PROPOSAL: struct-access-path aware TBAA

[Daniel Berlin]

On Tue, Mar 12, 2013 at 7:21 PM, Krzysztof Parzyszek
<kparzysz at codeaurora.org> wrote:
> On 3/12/2013 12:13 PM, Manman Ren wrote:
>>
>> Given
>>    struct A {
>>      int x;
>>      int y;
>>    };
>>    struct B {
>>      A a;
>>      int z;
>>    };
>>    struct C {
>>      B b1;
>>      B b2;
>>    };
>>    struct D {
>>      C c;
>>    };
>>
>> with struct-access-path aware TBAA, C::b1.a.x does not alias with
>> D::c.b2.a.x.
>> without it, the 2 scalar accesses can alias since both have int type.
>
> I browsed the 2012 standard for a while and I didn't see anything that would
> make this illegal:
>
> char *p = malloc(enough_bytes);
> intptr_t x = reinterpret_cast<intptr_t>(p);
> x += offsetof(C, b2);
> D &vd = *reinterpret_cast<D*>(p);
> C &vc = *reinterpret_cast<C*>(x);
> vd.c.b2.a.x = 1;    // ..accessing the same
> int t = vc.b1.a.x;  // ..storage
>
> I don't think that the path through the type structure is really sufficient.

There are simpler examples of this kind for  C++, because placement
new can change the dynamic type of the object (I actually haven't
looked to see if they changed this in 2012, but it was definitely
legal in C++98):

#include <new>
struct Foo { long i; };
struct Bar { void *p; };
long foo(int n)
{
  Foo *f = new Foo;
  f->i = 1;
  for (int i=0; i<n; ++i)
    {
      Bar *b = new (f) Bar;
      b->p = 0;
      f = new (f) Foo;
      f->i = i;
    }
  return f->i;
}

Both access to the same memory, both will end up with completely
different access paths, both legal by TBAA rules, but access path
alone will claim no-alias.

(This is taken from http://gcc.gnu.org/bugzilla/show_bug.cgi?id=29286)